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滑动窗口
给你两个长度相同的字符串，s 和 t。将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可,"> 
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        <h1 class="title">leetcode1208.尽可能使字符串相等</h1>
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            <span>二月 21, 2020</span>
            

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            <h1 id="leetcode1208-尽可能使字符串相等"><a href="#leetcode1208-尽可能使字符串相等" class="headerlink" title="leetcode1208. 尽可能使字符串相等"></a>leetcode1208. 尽可能使字符串相等</h1><hr>
<h3 id="滑动窗口"><a href="#滑动窗口" class="headerlink" title="滑动窗口"></a>滑动窗口</h3><blockquote>
<p>给你两个长度相同的字符串，s 和 t。<br>将 s 中的第 i 个字符变到 t 中的第 i 个字符需要 |s[i] - t[i]| 的开销（开销可能为 0），也就是两个字符的 ASCII 码值的差的绝对值。<br>用于变更字符串的最大预算是 maxCost。在转化字符串时，总开销应当小于等于该预算，这也意味着字符串的转化可能是不完全的。<br>如果你可以将 s 的子字符串转化为它在 t 中对应的子字符串，则返回可以转化的最大长度。<br>如果 s 中没有子字符串可以转化成 t 中对应的子字符串，则返回 0。</p>
</blockquote>
<blockquote>
<p>示例 1：<br>输入：s = “abcd”, t = “bcdf”, cost = 3<br>输出：3<br>解释：s 中的 “abc” 可以变为 “bcd”。开销为 3，所以最大长度为 3。</p>
</blockquote>
<blockquote>
<p>示例 2：<br>输入：s = “abcd”, t = “cdef”, cost = 3<br>输出：1<br>解释：s 中的任一字符要想变成 t 中对应的字符，其开销都是 2。因此，最大长度为 1。</p>
</blockquote>
<blockquote>
<p>示例 3：<br>输入：s = “abcd”, t = “acde”, cost = 0<br>输出：1<br>解释：你无法作出任何改动，所以最大长度为 1。</p>
</blockquote>
<p>首先，在我自己做的时候没看明白”krrgw” “zjxss” ，这组样例为什么是2，我开始认为krr-&gt;jxs，但是这道题是想让对应位置上做转化，也就是krr必须对应zjx。<br>这样就可以变成对应位置转化多少，才能不大于maxCount并且最长。</p>
<h3 id="也就是说：求一个数组的子数组，这个子数组的和不能大于maxCount。"><a href="#也就是说：求一个数组的子数组，这个子数组的和不能大于maxCount。" class="headerlink" title="也就是说：求一个数组的子数组，这个子数组的和不能大于maxCount。"></a>也就是说：求一个数组的子数组，这个子数组的和不能大于maxCount。</h3><p>属于滑动窗口的大小不固定的类型。右侧窗口不断扩大的时候记录长度，当窗口里面的数组和大于maxCount的时候，这个时候右窗口不能再扩大了，需要左边缩进来，直到满足数组和小于maxCount。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">equalSubstring</span><span class="params">(String s, String t, <span class="keyword">int</span> maxCost)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> result=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span>[] arr=<span class="keyword">new</span> <span class="keyword">int</span>[s.length()];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i&lt;arr.length;i++) &#123;</span><br><span class="line">        	arr[i]=Math.abs(t.charAt(i)-s.charAt(i));</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">int</span> sum=<span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>,i=<span class="number">0</span>;j&lt;t.length();j++) &#123;</span><br><span class="line">        	sum+=arr[j];</span><br><span class="line">        	<span class="keyword">while</span>(sum&gt;maxCost) &#123;</span><br><span class="line">        		sum-=arr[i];</span><br><span class="line">        		i++;</span><br><span class="line">        	&#125;</span><br><span class="line">        	result=Math.max(j-i+<span class="number">1</span>, result);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> result;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>

<p><strong>leetcode 6/100</strong></p>

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